Integration of $e^{ax}cos^n bx$ and $e^{ax}sin^n bx$












0















Q: Integration of $e^{ax}cos^n bx$ and $e^{ax}sin^n bx$




I know how to integration $e^{ax}cos bx$ using $cos bx=frac{e^{ibx}+e^{-ibx}}{2}$.Using the same trick here I got $$int e^{ax}cos^n bx ~dx=int e^{ax}left(frac{e^{ibx}+e^{-ibx}}{2}right)^n~ dx$$But it doesn't look easy to solve.I google it and find the result only which is:




$$int e^{ax}cos^n bx ~dx=frac{bnsin (bx)+acos (bx)}{a^2+b^2n^2}e^{ax}+frac{b^2(n-1)nint e^{ax}cos^{n-2}bxdx}{a^2+b^2n^2}.$$




Can Anyone help me to figure out this. Any hints or solution will be appreciated.

Thanks in advance.






complex-integration reduction-formula







share|cite|improve this question
























  • That "result" is a reduction formula, proved by integrating by parts.
    – Lord Shark the Unknown
    Nov 24 at 6:16










  • You can expand your formula by the binomial theorem and integrate termwise.
    – Lord Shark the Unknown
    Nov 24 at 6:17










  • If i use binomial theorem then it will consume a lot of time.Is there any other way to figure out this @Lord Shark the Unknown Sir
    – emonhossain
    Nov 24 at 6:18










  • As alreasy said, integrate by parts to get the reduction formula.
    – Claude Leibovici
    Nov 24 at 7:00










  • You may look at the method I used here.
    – Tianlalu
    Nov 24 at 7:09
















0















Q: Integration of $e^{ax}cos^n bx$ and $e^{ax}sin^n bx$




I know how to integration $e^{ax}cos bx$ using $cos bx=frac{e^{ibx}+e^{-ibx}}{2}$.Using the same trick here I got $$int e^{ax}cos^n bx ~dx=int e^{ax}left(frac{e^{ibx}+e^{-ibx}}{2}right)^n~ dx$$But it doesn't look easy to solve.I google it and find the result only which is:




$$int e^{ax}cos^n bx ~dx=frac{bnsin (bx)+acos (bx)}{a^2+b^2n^2}e^{ax}+frac{b^2(n-1)nint e^{ax}cos^{n-2}bxdx}{a^2+b^2n^2}.$$




Can Anyone help me to figure out this. Any hints or solution will be appreciated.

Thanks in advance.






complex-integration reduction-formula







share|cite|improve this question
























  • That "result" is a reduction formula, proved by integrating by parts.
    – Lord Shark the Unknown
    Nov 24 at 6:16










  • You can expand your formula by the binomial theorem and integrate termwise.
    – Lord Shark the Unknown
    Nov 24 at 6:17










  • If i use binomial theorem then it will consume a lot of time.Is there any other way to figure out this @Lord Shark the Unknown Sir
    – emonhossain
    Nov 24 at 6:18










  • As alreasy said, integrate by parts to get the reduction formula.
    – Claude Leibovici
    Nov 24 at 7:00










  • You may look at the method I used here.
    – Tianlalu
    Nov 24 at 7:09














0












0








0


2






Q: Integration of $e^{ax}cos^n bx$ and $e^{ax}sin^n bx$




I know how to integration $e^{ax}cos bx$ using $cos bx=frac{e^{ibx}+e^{-ibx}}{2}$.Using the same trick here I got $$int e^{ax}cos^n bx ~dx=int e^{ax}left(frac{e^{ibx}+e^{-ibx}}{2}right)^n~ dx$$But it doesn't look easy to solve.I google it and find the result only which is:




$$int e^{ax}cos^n bx ~dx=frac{bnsin (bx)+acos (bx)}{a^2+b^2n^2}e^{ax}+frac{b^2(n-1)nint e^{ax}cos^{n-2}bxdx}{a^2+b^2n^2}.$$




Can Anyone help me to figure out this. Any hints or solution will be appreciated.

Thanks in advance.






complex-integration reduction-formula







share|cite|improve this question
















Q: Integration of $e^{ax}cos^n bx$ and $e^{ax}sin^n bx$




I know how to integration $e^{ax}cos bx$ using $cos bx=frac{e^{ibx}+e^{-ibx}}{2}$.Using the same trick here I got $$int e^{ax}cos^n bx ~dx=int e^{ax}left(frac{e^{ibx}+e^{-ibx}}{2}right)^n~ dx$$But it doesn't look easy to solve.I google it and find the result only which is:




$$int e^{ax}cos^n bx ~dx=frac{bnsin (bx)+acos (bx)}{a^2+b^2n^2}e^{ax}+frac{b^2(n-1)nint e^{ax}cos^{n-2}bxdx}{a^2+b^2n^2}.$$




Can Anyone help me to figure out this. Any hints or solution will be appreciated.

Thanks in advance.






complex-integration reduction-formula




complex-integration reduction-formula






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 at 4:19









clathratus

3,017330




3,017330










asked Nov 24 at 6:14









emonhossain

33611




33611












  • That "result" is a reduction formula, proved by integrating by parts.
    – Lord Shark the Unknown
    Nov 24 at 6:16










  • You can expand your formula by the binomial theorem and integrate termwise.
    – Lord Shark the Unknown
    Nov 24 at 6:17










  • If i use binomial theorem then it will consume a lot of time.Is there any other way to figure out this @Lord Shark the Unknown Sir
    – emonhossain
    Nov 24 at 6:18










  • As alreasy said, integrate by parts to get the reduction formula.
    – Claude Leibovici
    Nov 24 at 7:00










  • You may look at the method I used here.
    – Tianlalu
    Nov 24 at 7:09


















  • That "result" is a reduction formula, proved by integrating by parts.
    – Lord Shark the Unknown
    Nov 24 at 6:16










  • You can expand your formula by the binomial theorem and integrate termwise.
    – Lord Shark the Unknown
    Nov 24 at 6:17










  • If i use binomial theorem then it will consume a lot of time.Is there any other way to figure out this @Lord Shark the Unknown Sir
    – emonhossain
    Nov 24 at 6:18










  • As alreasy said, integrate by parts to get the reduction formula.
    – Claude Leibovici
    Nov 24 at 7:00










  • You may look at the method I used here.
    – Tianlalu
    Nov 24 at 7:09
















That "result" is a reduction formula, proved by integrating by parts.
– Lord Shark the Unknown
Nov 24 at 6:16




That "result" is a reduction formula, proved by integrating by parts.
– Lord Shark the Unknown
Nov 24 at 6:16












You can expand your formula by the binomial theorem and integrate termwise.
– Lord Shark the Unknown
Nov 24 at 6:17




You can expand your formula by the binomial theorem and integrate termwise.
– Lord Shark the Unknown
Nov 24 at 6:17












If i use binomial theorem then it will consume a lot of time.Is there any other way to figure out this @Lord Shark the Unknown Sir
– emonhossain
Nov 24 at 6:18




If i use binomial theorem then it will consume a lot of time.Is there any other way to figure out this @Lord Shark the Unknown Sir
– emonhossain
Nov 24 at 6:18












As alreasy said, integrate by parts to get the reduction formula.
– Claude Leibovici
Nov 24 at 7:00




As alreasy said, integrate by parts to get the reduction formula.
– Claude Leibovici
Nov 24 at 7:00












You may look at the method I used here.
– Tianlalu
Nov 24 at 7:09




You may look at the method I used here.
– Tianlalu
Nov 24 at 7:09










1 Answer
1






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oldest

votes


















1














(Differentiation is easier than integration)



Let $f_{m,a, b}(x)=f_m=e^{ax}cos^m bx$. Take the derivative of $f_{m}$,
begin{align*}
f'_{m}&=af_{m}-bmf_{m-1}sin bx.tag{1}
end{align*}
So, if we take $g_{m}=af_{m}color{magenta}+bmf_{m-1}sin bx$, the derivative of $g_{m}$ will cancel $color{blue}{sin bx}$ out, i.e.
begin{align*}
g'_{m}&=af'_{m}hspace{8.2em}+bmbig(f'_{m-1}sin bxhspace{10.8em}+bunderbrace{f_{m-1}cos bx}_{=f_{m}}big)\
g'_{m}&=a^2f_{m}-color{blue}{abmf_{m-1}sin bx}+color{blue}{bm}big(color{blue}{af_{m-1}sin bx}-b(m-1) f_{m-2}color{orange}{sin^2 bx}+bf_mbig)\
g'_{m}&=(a^2+b^2m)f_m-b^2m(m-1)f_{m-2}color{orange}{(1-cos^2 bx)}\
g'_{m}&=(a^2+b^2m^2)f_m-b^2m(m-1)f_{m-2}.tag{2}
end{align*}
Now, we assume
$$I_{m}=int f_m~mathrm dx=int e^{ax}cos^m bx~mathrm dx,$$
integrate $(2)$, we get the desired result




$$g_m=(a^2+b^2m^2)I_{m}-b^2m(m-1)I_{m-2}.$$







share|cite|improve this answer





















  • Thanks @Tianlalu Really appreciated.Can i use Differentiation to get Integration in this way always?
    – emonhossain
    Nov 24 at 9:05






  • 1




    Yes, we can always do so once we find the appropriate $f_m$. For example, for $int cos^n x~dx$, we may choose $f_n=cos^{n-1}x sin x$ and differentiate it to cancel $sin x$ out.
    – Tianlalu
    Nov 24 at 9:11













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1 Answer
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1














(Differentiation is easier than integration)



Let $f_{m,a, b}(x)=f_m=e^{ax}cos^m bx$. Take the derivative of $f_{m}$,
begin{align*}
f'_{m}&=af_{m}-bmf_{m-1}sin bx.tag{1}
end{align*}
So, if we take $g_{m}=af_{m}color{magenta}+bmf_{m-1}sin bx$, the derivative of $g_{m}$ will cancel $color{blue}{sin bx}$ out, i.e.
begin{align*}
g'_{m}&=af'_{m}hspace{8.2em}+bmbig(f'_{m-1}sin bxhspace{10.8em}+bunderbrace{f_{m-1}cos bx}_{=f_{m}}big)\
g'_{m}&=a^2f_{m}-color{blue}{abmf_{m-1}sin bx}+color{blue}{bm}big(color{blue}{af_{m-1}sin bx}-b(m-1) f_{m-2}color{orange}{sin^2 bx}+bf_mbig)\
g'_{m}&=(a^2+b^2m)f_m-b^2m(m-1)f_{m-2}color{orange}{(1-cos^2 bx)}\
g'_{m}&=(a^2+b^2m^2)f_m-b^2m(m-1)f_{m-2}.tag{2}
end{align*}
Now, we assume
$$I_{m}=int f_m~mathrm dx=int e^{ax}cos^m bx~mathrm dx,$$
integrate $(2)$, we get the desired result




$$g_m=(a^2+b^2m^2)I_{m}-b^2m(m-1)I_{m-2}.$$







share|cite|improve this answer





















  • Thanks @Tianlalu Really appreciated.Can i use Differentiation to get Integration in this way always?
    – emonhossain
    Nov 24 at 9:05






  • 1




    Yes, we can always do so once we find the appropriate $f_m$. For example, for $int cos^n x~dx$, we may choose $f_n=cos^{n-1}x sin x$ and differentiate it to cancel $sin x$ out.
    – Tianlalu
    Nov 24 at 9:11


















1














(Differentiation is easier than integration)



Let $f_{m,a, b}(x)=f_m=e^{ax}cos^m bx$. Take the derivative of $f_{m}$,
begin{align*}
f'_{m}&=af_{m}-bmf_{m-1}sin bx.tag{1}
end{align*}
So, if we take $g_{m}=af_{m}color{magenta}+bmf_{m-1}sin bx$, the derivative of $g_{m}$ will cancel $color{blue}{sin bx}$ out, i.e.
begin{align*}
g'_{m}&=af'_{m}hspace{8.2em}+bmbig(f'_{m-1}sin bxhspace{10.8em}+bunderbrace{f_{m-1}cos bx}_{=f_{m}}big)\
g'_{m}&=a^2f_{m}-color{blue}{abmf_{m-1}sin bx}+color{blue}{bm}big(color{blue}{af_{m-1}sin bx}-b(m-1) f_{m-2}color{orange}{sin^2 bx}+bf_mbig)\
g'_{m}&=(a^2+b^2m)f_m-b^2m(m-1)f_{m-2}color{orange}{(1-cos^2 bx)}\
g'_{m}&=(a^2+b^2m^2)f_m-b^2m(m-1)f_{m-2}.tag{2}
end{align*}
Now, we assume
$$I_{m}=int f_m~mathrm dx=int e^{ax}cos^m bx~mathrm dx,$$
integrate $(2)$, we get the desired result




$$g_m=(a^2+b^2m^2)I_{m}-b^2m(m-1)I_{m-2}.$$







share|cite|improve this answer





















  • Thanks @Tianlalu Really appreciated.Can i use Differentiation to get Integration in this way always?
    – emonhossain
    Nov 24 at 9:05






  • 1




    Yes, we can always do so once we find the appropriate $f_m$. For example, for $int cos^n x~dx$, we may choose $f_n=cos^{n-1}x sin x$ and differentiate it to cancel $sin x$ out.
    – Tianlalu
    Nov 24 at 9:11
















1












1








1






(Differentiation is easier than integration)



Let $f_{m,a, b}(x)=f_m=e^{ax}cos^m bx$. Take the derivative of $f_{m}$,
begin{align*}
f'_{m}&=af_{m}-bmf_{m-1}sin bx.tag{1}
end{align*}
So, if we take $g_{m}=af_{m}color{magenta}+bmf_{m-1}sin bx$, the derivative of $g_{m}$ will cancel $color{blue}{sin bx}$ out, i.e.
begin{align*}
g'_{m}&=af'_{m}hspace{8.2em}+bmbig(f'_{m-1}sin bxhspace{10.8em}+bunderbrace{f_{m-1}cos bx}_{=f_{m}}big)\
g'_{m}&=a^2f_{m}-color{blue}{abmf_{m-1}sin bx}+color{blue}{bm}big(color{blue}{af_{m-1}sin bx}-b(m-1) f_{m-2}color{orange}{sin^2 bx}+bf_mbig)\
g'_{m}&=(a^2+b^2m)f_m-b^2m(m-1)f_{m-2}color{orange}{(1-cos^2 bx)}\
g'_{m}&=(a^2+b^2m^2)f_m-b^2m(m-1)f_{m-2}.tag{2}
end{align*}
Now, we assume
$$I_{m}=int f_m~mathrm dx=int e^{ax}cos^m bx~mathrm dx,$$
integrate $(2)$, we get the desired result




$$g_m=(a^2+b^2m^2)I_{m}-b^2m(m-1)I_{m-2}.$$







share|cite|improve this answer












(Differentiation is easier than integration)



Let $f_{m,a, b}(x)=f_m=e^{ax}cos^m bx$. Take the derivative of $f_{m}$,
begin{align*}
f'_{m}&=af_{m}-bmf_{m-1}sin bx.tag{1}
end{align*}
So, if we take $g_{m}=af_{m}color{magenta}+bmf_{m-1}sin bx$, the derivative of $g_{m}$ will cancel $color{blue}{sin bx}$ out, i.e.
begin{align*}
g'_{m}&=af'_{m}hspace{8.2em}+bmbig(f'_{m-1}sin bxhspace{10.8em}+bunderbrace{f_{m-1}cos bx}_{=f_{m}}big)\
g'_{m}&=a^2f_{m}-color{blue}{abmf_{m-1}sin bx}+color{blue}{bm}big(color{blue}{af_{m-1}sin bx}-b(m-1) f_{m-2}color{orange}{sin^2 bx}+bf_mbig)\
g'_{m}&=(a^2+b^2m)f_m-b^2m(m-1)f_{m-2}color{orange}{(1-cos^2 bx)}\
g'_{m}&=(a^2+b^2m^2)f_m-b^2m(m-1)f_{m-2}.tag{2}
end{align*}
Now, we assume
$$I_{m}=int f_m~mathrm dx=int e^{ax}cos^m bx~mathrm dx,$$
integrate $(2)$, we get the desired result




$$g_m=(a^2+b^2m^2)I_{m}-b^2m(m-1)I_{m-2}.$$








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 24 at 8:42









Tianlalu

3,12321038




3,12321038












  • Thanks @Tianlalu Really appreciated.Can i use Differentiation to get Integration in this way always?
    – emonhossain
    Nov 24 at 9:05






  • 1




    Yes, we can always do so once we find the appropriate $f_m$. For example, for $int cos^n x~dx$, we may choose $f_n=cos^{n-1}x sin x$ and differentiate it to cancel $sin x$ out.
    – Tianlalu
    Nov 24 at 9:11




















  • Thanks @Tianlalu Really appreciated.Can i use Differentiation to get Integration in this way always?
    – emonhossain
    Nov 24 at 9:05






  • 1




    Yes, we can always do so once we find the appropriate $f_m$. For example, for $int cos^n x~dx$, we may choose $f_n=cos^{n-1}x sin x$ and differentiate it to cancel $sin x$ out.
    – Tianlalu
    Nov 24 at 9:11


















Thanks @Tianlalu Really appreciated.Can i use Differentiation to get Integration in this way always?
– emonhossain
Nov 24 at 9:05




Thanks @Tianlalu Really appreciated.Can i use Differentiation to get Integration in this way always?
– emonhossain
Nov 24 at 9:05




1




1




Yes, we can always do so once we find the appropriate $f_m$. For example, for $int cos^n x~dx$, we may choose $f_n=cos^{n-1}x sin x$ and differentiate it to cancel $sin x$ out.
– Tianlalu
Nov 24 at 9:11






Yes, we can always do so once we find the appropriate $f_m$. For example, for $int cos^n x~dx$, we may choose $f_n=cos^{n-1}x sin x$ and differentiate it to cancel $sin x$ out.
– Tianlalu
Nov 24 at 9:11




















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